Given the polynomial P(x) = 2x3 − 5x2 − 19x + 42, find the quotient and remainder when P(x) is divided by 2x-7.

Here's how to solve these polynomial problems.

Given the polynomial $P(x) = 2x^3 - 5x^2 - 19x + 42$.

a) Find the quotient and the remainder for the division of $P(x)$ by $2x-7$.

We will use polynomial long division.

Step 1: Set up the long division.

$$\begin{array}{r} x^2 + x - 6 \\[-3pt] 2x-7 \overline{\smash) 2x^3 - 5x^2 - 19x + 42} \\[-3pt] \end{array}$$

Step 2: Divide the leading terms.
Divide $2x^3$ by $2x$ to get $x^2$. Multiply $x^2$ by $(2x-7)$ and subtract.

$$\begin{array}{r} x^2 \phantom{+x-6} \\[-3pt] 2x-7 \overline{\smash) 2x^3 - 5x^2 - 19x + 42} \\[-3pt] -(2x^3 - 7x^2) \\[-3pt] \hline 2x^2 - 19x \\[-3pt] \end{array}$$

Step 3: Repeat the process.
Bring down the next term ($-19x$). Divide $2x^2$ by $2x$ to get $x$. Multiply $x$ by $(2x-7)$ and subtract.

$$\begin{array}{r} x^2 + x \phantom{-6} \\[-3pt] 2x-7 \overline{\smash) 2x^3 - 5x^2 - 19x + 42} \\[-3pt] -(2x^3 - 7x^2) \\[-3pt] \hline 2x^2 - 19x \\[-3pt] -(2x^2 - 7x) \\[-3pt] \hline -12x + 42 \\[-3pt] \end{array}$$

Step 4: Repeat again.
Bring down the next term ($+42$). Divide $-12x$ by $2x$ to get $-6$. Multiply $-6$ by $(2x-7)$ and subtract.

$$\begin{array}{r} x^2 + x - 6 \\[-3pt] 2x-7 \overline{\smash) 2x^3 - 5x^2 - 19x + 42} \\[-3pt] -(2x^3 - 7x^2) \\[-3pt] \hline 2x^2 - 19x \\[-3pt] -(2x^2 - 7x) \\[-3pt] \hline -12x + 42 \\[-3pt] -(-12x + 42) \\[-3pt] \hline 0 \\[-3pt] \end{array}$$

The quotient is $x^2 + x - 6$ and the remainder is 0.

The quotient is $\boxed{x^2 + x - 6}$ and the remainder is $\boxed{0}$.

b) Factorise $P(x)$ completely.

Since the remainder from part (a) is 0, $(2x-7)$ is a factor of $P(x)$.
So, $P(x) = (2x-7)(x^2 + x - 6)$.

Step 1: Factor the quadratic term.
We need to factor the quadratic expression $x^2 + x - 6$. We look for two numbers that multiply to $-6$ and add to 1. These numbers are 3 and $-2$.
So, $x^2 + x - 6 = (x+3)(x-2)$.

Step 2: Write the completely factored form.
Substitute the factored quadratic back into the expression for $P(x)$.

$$P(x) = (2x-7)(x+3)(x-2)$$

The completely factored form is $\boxed{(2x-7)(x+3)(x-2)}$.

c) Solve if $P(x) = 0$.

To solve $P(x) = 0$, we set the factored form of $P(x)$ equal to zero.

$$(2x-7)(x+3)(x-2) = 0$$

Step 1: Set each factor to zero and solve for $x$.

• For the first factor: $$2x-7 = 0$$ $$2x = 7$$ $$x = \frac{7}{2}$$
• For the second factor: $$x+3 = 0$$ $$x = -3$$
• For the third factor: $$x-2 = 0$$ $$x = 2$$

The solutions are $\boxed{x = \frac{7}{2}, x = -3, x = 2}$.

d) Solve: $2x^3 + 9x^2 - 2x - 24 = 0$.

Let $Q(x) = 2x^3 + 9x^2 - 2x - 24$. We need to find the roots of this cubic equation.

Step 1: Find a rational root using the Rational Root Theorem or by testing values.
We test integer factors of the constant term $(-24)$ divided by integer factors of the leading coefficient $(2)$.
Let's try $x = -2$:

$$Q(-2) = 2(-2)^3 + 9(-2)^2 - 2(-2) - 24$$ $$Q(-2) = 2(-8) + 9(4) + 4 - 24$$ $$Q(-2) = -16 + 36 + 4 - 24$$ $$Q(-2) = 40 - 40$$ $$Q(-2) = 0$$

Since $Q(-2) = 0$, $x = -2$ is a root, and $(x+2)$ is a factor of $Q(x)$.

Step 2: Use synthetic division to divide $Q(x)$ by $(x+2)$.
The coefficients of $Q(x)$ are $2, 9, -2, -24$.

$$\begin{array}{c|cccc} -2 & 2 & 9 & -2 & -24 \\ & & -4 & -10 & 24 \\ \hline & 2 & 5 & -12 & 0 \\ \end{array}$$

The quotient is $2x^2 + 5x - 12$.
So, $Q(x) = (x+2)(2x^2 + 5x - 12) = 0$.

Step 3: Factor the quadratic quotient.
We need to solve $2x^2 + 5x - 12 = 0$. We can factor this quadratic.
We look for two numbers that multiply to $2 \times -12 = -24$ and add to 5. These numbers are 8 and $-3$.

$$2x^2 + 8x - 3x - 12 = 0$$ $$2x(x+4) - 3(x+4) = 0$$ $$(2x-3)(x+4) = 0$$

Step 4: Set each factor to zero and solve for $x$.

• From the first factor we found: $$x+2 = 0$$ $$x = -2$$
• From the quadratic factors: $$2x-3 = 0$$ $$2x = 3$$ $$x = \frac{3}{2}$$ $$x+4 = 0$$ $$x = -4$$

The solutions are $\boxed{x = -2, x = \frac{3}{2}, x = -4}$.